Problem: Amelia flies her airplane through calm skies (no wind) at a velocity (speed and direction) $\vec{v_1}$. The direction of $\vec{v_1}$ is $15^\circ$ north of east, and the speed is $180\,\text{km/h}$. Eventually, however, she enters a windy part of the atmosphere and finds that her plane now moves at a velocity $\vec{v_2}$. The direction of $\vec{v_2}$ is due east, and the speed is $150\,\text{km/h}$. (Assume "due east" is $0^\circ$, "due north" is $90^\circ$, and so on.) What is the speed of the wind?
Answer: Consider vector $\vec w$ (depicted below), which represents the wind. We can imagine how it would cause Amelia's airplane to slow down and change direction. It is reasonable to assume that the velocity of the wind added with the initial velocity of Amelia's airplane equals the resultant velocity of her airplane. $\vec w + \vec{v_1} = \vec{v_2}$ Before finding $\vec w$, we'll need to split both vectors into their components. $\vec{v_1}$ can be broken into its components using the sine and cosine functions. (Angle not drawn to scale) $x = 180\cos{(15^\circ)} \approx 173.867$ $y = 180\sin{(15^\circ)} = 46.587$ Therefore, $\vec{v_1} = (173.867,\, 46.587).$ $\vec{v_2}$ is $150\,\text{km/h}$ due east, which can be written as $(150, 0)$. We can now solve for $\vec w$. $\begin{aligned} \vec w + \vec{v_1} &= \vec{v_2}\\\\ \vec w &= \vec{v_2} - \vec{v_1}\\\\ \vec w &= (150, 0) - (173.867,\, 46.587)\\\\ \vec w &= (-23.867, \,-46.587) \end{aligned}$ We can find the magnitude of $\vec w$ (i.e., the speed of the wind) using the Pythagorean theorem. $\begin{aligned} \| \vec w \|^2 &= (-23.867)^2 + (-46.587)^2\\\\ \| \vec w \| &= \sqrt{(-23.867)^2 + (-46.587)^2}\\\\ \| \vec w \| &\approx 52.3 \text{ km/h (rounded to the nearest tenth)} \end{aligned}$ $\vec w$ is pointing into the third quadrant, so we can find its direction (call it $\theta~$ ) using the arctangent function and adding $180^\circ$. $\begin{aligned} \tan \theta &= \dfrac{-46.587}{-23.867}\\\\ \theta &= \arctan{ \left ( \dfrac{46.587}{23.867} \right ) } \\\\ \theta&\approx 63^\circ \end{aligned}$ Adding $180^\circ$ to this result gives us the actual direction, $243^\circ$ (rounded to the nearest degree). The speed of the wind is $52.3 \text{ km}/\text{h}$. The direction of the wind is $243^\circ$.